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Hey all-just a student trying to figure out what's going on. Don't be a stranger :D!


Dec
11
answered Intersection between a closed set and $y=x$ on $[0,1]$
Dec
5
comment construct non-commutative group with prescribed center
Why not take the product with $PGL(2)$?
Dec
5
answered The set of all fixed points of a continuous function $f:[0,1] \to [0,1]$ , satisfying $f \circ f=f$ , is a non-empty interval ?
Dec
5
comment Probability of hitting a point in a square, maximum of coordinates
Draw some level sets for the function $\max(x,y)$ on $[0,1]$ - you'll get a bunch of $L$'s. Then it will be clear how to calculate the area.
Dec
5
comment Flatness on the affine line for a coherent sheaf
@user54369 Nothing to apologize for, but still not true: take something like the trivial line bundle and throw in some junk at a few points (so $M = \mathbb{C}[t] \oplus \mathbb{C}[t]/t$, say) - then at every point except the origin, $M$ looks nice and free, but at the origin it's all messed up.
Nov
2
comment Matrix of Shift Transform on arbitrary basis
You are almost there - you have every right to think of $v_1, \ldots, v_n$ as the `standard' basis, so your matrix looks like the identity matrix shifted down one row in all but the last column, where it reads as you wrote.
Oct
23
comment set of all regular values
@SarahT. you are absolutely right, I was being careless. Good catch!
Oct
22
comment set of all regular values
As before it suffices to check on a cover: now take a cover $U_i$ for $N$ with each $U_i \simeq \mathbb{R}^n$, and take a cover $V_{ij}$ for $f^{-1}(U_i)$ with each $V_{ij} \simeq \mathbb{R}^m$, with this it suffices to check that for a map $\mathbb{R}^m \rightarrow \mathbb{R}^n$, the noncritical locus is open. At a point in it $x$, the Jacobian matrix $df$ has full rank - it is a linear algebra exercise that this means some $\dim N$ minor of it has full rank - since this minor's determinant is continuous as a function on $\mathbb{R}^m$, it is still nonzero nearby, so still $df$ is full rank.
Oct
22
comment set of all regular values
Yes, you're welcome. Now use that at $x$, some $\dim N$ minor of the Jacobian has nonvanishing determinant (this works for $\geqslant$ as well).
Oct
22
comment set of all regular values
Okay: you can have an open cover of $M$ by copies $U_i$ of $\mathbb{R}^n$, and you can check that a set $W$ is open iff $W \cap U_i$ for all $i$ - this is just point set topology. With that, you're reduced to answering the question for $\mathbb{R}^n$ - you want say that if $x$ is not a critical value, nearby guys are not either. What does it mean to be critical? All your partials vanish. So at $x$, some $\frac{\partial f}{\partial x_i} \neq 0$, by continuity this is true nearby.
Oct
22
answered set of all regular values
Oct
22
answered Confused about short exact sequence involving $\mathcal{O}_{\mathbb{P}^n}$ and $\mathcal{O}_Z$ for $\pi:Z\rightarrow \mathbb{P}^n$ closed embedding
Sep
24
awarded  Autobiographer
Aug
24
answered Definition of a coordinate vector bundle
Aug
24
comment Definition of a coordinate vector bundle
By 1., certainly each fibre of $B$ can be set-theoretically identified with $\mathbb{R}^n$ over $U_f$, using $\phi_f$. By 2., the induced vector space structure doesn't depend on your choice of $f$.
Aug
24
comment Convergence in measure implies pointwise convergence?
@mathematician That seems true to me - if a sequence $a_i$ doesn't converge to $a$, we can find a subsequence which is always at least $\epsilon$ away from $a$, which is preserved by any subsequence. Perhaps the problem is that you are only given a subsequence which converges almost everywhere, and there are uncountably many such sequences (so the total problem area where we don't converge pointwise need not have measure 0)? (convergence in measure is definitely preserved by taking subsequences)
Apr
15
comment The definition of the quotient category in abelian category.
Hi, maybe a silly point, but in case anyone else reads this and is confused: I didn't understand how $A/(A' \cap A'')$ is a quotient of $A/A'$; however, it does fit in the SES $$0 \rightarrow A''/(A' \cap A'') \rightarrow A/(A' \cap A'') \rightarrow A/A'' \rightarrow 0$$where the left and right lie in $C$.
Apr
12
comment differential forms, stokes theorem in higher dimension
Glad to help - I also found differential forms challenging when learning (and relearning, and relearning...) them, and had the same experience re: few concrete calculations. Best wishes
Apr
11
comment measure space problem.
Hi, I've added some more details, let me know if you still have questions. You're welcome :)
Apr
11
revised measure space problem.
added 1065 characters in body